Let $n \in \mathbb{N}^*$ and $x_1,x_2,x_3,\ldots,x_n \in \mathbb{R}$' such that $0\le x_1\le x_2 \le \cdots \le x_n$, and $x_1+x_2+x_3+\cdots+x_n=1$, show that: $$(1+x_1^21^2)(1+x_2^22^2)\cdots(1+x_n^2n^2)\ge \frac{2n^2+9n+1}{6n}$$
I have no idea how you could solve this. I tried to manipulate the inequality using the inequality of arithmetic and geometric mean but it led me nowhere. I'm really interested in the solution of this problem. Any ideas?
Consider two adjacent terms in the product, and use $x=x_j,y=x_{j+1}$ for the notation. The nondecreasing inequality then gives the constraint $x \le y$ and there is a positive integer $a=j^2$ so that the contribution to the product from the $j$ and $j+1$ terms is $$(1+a^2x^2)(1+(a+1)^2y^2). \tag{1}$$ Now if initially $x+y=k$ we only decrease (or keep the same) this two term product by replacing $x,y$ each by $k/2$, and note that since initially $x \le y$ this replacement has increased (or kept the same) $x$ while decreasing (or keeping the same) $y$, and so the replacement is consistent with the nondecreasing inequality among the $x_j$'s. It is also clear that, if $x\neq y$ were true initially, the replaced contribution would be strictly less than the initial product $(1).$
This argument shows that (as Greg Martin suggested) the minimal value of the expression occurs when all the $x_j$ are $1/n.$ If $p(n)=(2n^2+9n+1)/(6n)$ and $q(n)$ is the value of the product when each $x_j=1/n,$ then $p(1)=q(1)=2,$ so that for general $n$ the minimal lower bound of $p(n)$ happens to be exact at $n=1$. For larger $n$ it is less than $q(n)$, and has to be checked by hand for $n \le 4$, for which $q(2)-p(2)=1/4,$ and $q(3)-p(3)=53/81,$ and finally $q(4)-p(4)=655/512.$ [There may be a more clever way to see in general that $q(n)>p(n)$ but I don't see it.]
As soon as $n \ge 5$ we can use the fact that $1/n$ times the logarithm of the product is a right-endpoint sum for the integral $$\int_0^1 \log(1+x^2)\ dx = c= \log 2+(\pi-4)/2.$$ Then from $(1/n)\cdot \log q(n) \ge c$ we get $q(n) \ge e^{cn}.$ Though I didn't do this last part carefully, it seems that for $n \ge 5$ we do have $e^{cn} >p(n)$ as required. The left side here should certainly overtake the right since it's exponential with $e^c \approx 1.302.$ I guess the reason the integral bound isn't tight for $n<5$ is because too much error occurs in replacing the sum by the integral.
EDIT (at suggestion by the OP BarbuDorel in a comment). Once it is agreed the minimum of the product occurs when all the $x_j$ are $1/n$ the product becomes $$(1+(1/n)^2)(1+(2/n)^2)\cdots (1+(n/n)^2).$$ This is clearly at least $1$ plus the sum of the squared terms which occur as second terms in the factors. But $$1+(1/n)^2+(2/n)^2+\cdots+(n/n)^2=\frac{2n^2+9n+1}{6n},$$ on applying the formula for the sum of the first $n$ squares. So there is no need to consider integrals as I did above. [There still remains some work to justify replacing the product of two adjacent factors $(1)$ by the product obtained on replacing adjacent $x_j$ by their average. I'll add that if I can get a simple argument.
A simpler approach. Note that the given product is bounded below by the sum $$S=1+x_1^2+2^2x_2+\cdots + n^2x_n^2.$$ As noted above, if we can show this sum is minimal when all the $x_j$ are equal, the inequality follows. The argument for this goes similarly to before, looking at two adjacent terms, but the math is easier. Suppose the adjacent terms contribute $ax^2+by^2$ [where $0<a<b$] to the sum, and that initially $x<y$. Then if initially $x+y=k$ we have $x<k/2$. Now consider $$ax^2+b(k-x)^2=(a+b)x^2 -2bkx+bk^2.$$ The graph of this is a parabola opening upward, with its vertex at $x=bk/(a+b)>k/2,$ where the latter follows from $0<a<b.$ So since our initial $x$ satisfies $x<k/2$ we see that the value of $ax^2+by^2$ is strictly decreased when we replace each of $x,y$ by their average $(x+y)/2.$
Now since it is clear that the restrictions on the $x_k$ define a compact bounded region in $n$ space, the function $S$ must have a minimum, and the above argument shows this minimum cannot occur at any point at which any two of the $x_j$ are not equal. Conclusion: at the minimal $S$ all the $x_j$ are equal, and the overall inequality follows from that as just outlined.
